[Kiki240240]

Hello everyone. I am stuck on a question of determining an Ecell value. The information given is the cell notation:
Ni(s) l Ni(OH)2 (s) l OH- (0.022 M) ll Ni 2+ (0.16 M) l Ni(s)
As well as the Ksp of Ni(OH)2= 5.5x10^-16
So, I wrote out the anode and cathode equations like so:
Cathode: Ni2+ + 2e- ---> Ni
Anode: Ni + 2OH- ---> Ni(OH)2 + 2e-
Then I used the equation E°= (RT)/(nF)lnK
(8.314x298.15)/(2x96487)ln(1/5.5x10^-16)
And got 0.45V
This would be the E° of the anode. So, I think I need the E° of the cathode to use E°cell=E°cathode - E°anode
Then I would be able to use the Nernst equation Ecell=E°cell-(RT)/(nF)lnQ, using (0.16/0.022^2) as my Q.
Is any of my thought process correct? How would I find the E° of the cathode? Thank you for any insight you have to give!

[Borek]

I would calculate concentration of Ni²⁺ on the left from Ksp, and treat the problem as if it were the concentration cell.

[Big-Daddy]

All of your thought process is correct. You cannot find E° of the cathode or E° of the anode individually using the given information. You can only note a "coincidence":

Your RHS cell reaction is Ni²⁺(aq) + 2e^-  Ni(s)
Your LHS cell reaction is Ni(s) + 2OH^-(aq)  Ni(OH)₂(s) + 2e^-

The total cell reaction is thus Ni²⁺(aq) + 2OH^-(aq) Ni(OH)₂(s), n=2, from addition of the equations to balance e^-.

Now this cell reaction happens to be one for which you know ΔG°, because you happen to know K for solubility reaction Ni(OH)₂(s) Ni²⁺(aq) + 2OH^-(aq). The process occurring here is different (one is a dissolution, one is a redox exchange of electrons) but the path-invariance law for thermodynamic state functions ("Hess law") tells us that they will have the same ΔG and thus ΔG°. So you can convert K to ΔG° for the solubility reaction, equate that to ΔG° for the cell reaction, and from ΔG° for the cell reaction find E° for the cell reaction.

E = E° - RT/(nF) * ln(Q) will take it from there to find E of the cell given your activity or concentration values

You cannot find E° of the individual half-cells from this because you do not really have electrochemical information with which to tell such things. You only have a situation where, by the precisely managed nature of the two cells used, the cell reaction is equivalent to one for which you do know the equilibrium constant.

Since you asked for any insights, I would add that this all assumes [Ni²⁺][OH^-]² = K_sp. If [Ni²⁺][OH^-]² < K_sp then the solubility product equilibrium does not become established at all, and all the nickel hydroxide dissolves. Of course, you know it has been established because the cell set-up notes the presence of solid nickel hydroxide.

[Big-Daddy]

I would calculate concentration of Ni²⁺ on the left from Ksp, and treat the problem as if it were the concentration cell.

I don't think the concentration of Ni²⁺ on the LHS matters here does it?

The cell reaction is using Ni²⁺ from the RHS cell, so I'd think that's the only concentration from the initial-set up that you actually need? Ni²⁺ does not come into the half-reaction of the LHS AFAIK. (probably because its concentration is so low, ≈1.14·10^-12 M, on LHS, which is why we are ok with this cell reaction)

As for OH^-, it comes in on the LHS, hence the need to know the LHS cell's concentration of OH^-.

Don't think K_sp plays any real role here except that it happens to match the cell reaction equation

[Borek]

Don't think K_sp plays any real role here

Yes it does. Even you have suggested to use it in calculations.

You are right it is impossible to calculate potentials for cathode and anode separately, as they depend on the E₀ for Ni/Ni²⁺. But it doesn't matter, as in the concentration cell E₀ cancels out.

[Big-Daddy]

Even you have suggested to use it in calculations.

I meant that, beyond the equivalence of the reaction equation for which we have equilibrium constant K (=K_sp) to the cell reaction for our cell here, we do not need that solubility equilibrium.

To be precise, I just meant that you don't need to calculate Ni²⁺ concentration for the LHS cell or OH^- concentration for the RHS cell if you don't want to. (Cannot even do the latter calculation for sure, since we don't know if there is Ni(OH)2(s) precipitating on the RHS or not.) The potential, E, can be determined in terms of the concentrations given directly, because the cell reaction is really

Ni²⁺(aq,RHS) + 2OH^-(aq,LHS) Ni(OH)2(s,LHS), n=2

and we have both of these concentrations given in the OP.

[Borek]

The potential, E, can be determined in terms of the concentrations given directly, because the cell reaction is really

Ni²⁺(aq,RHS) + 2OH^-(aq,LHS) Ni(OH)2(s,LHS), n=2

and we have both of these concentrations given in the OP.

Why do you assume n=2 if the reaction doesn't involve charge transfer?

And what are you going to use for E0?

[Big-Daddy]

Why do you assume n=2 if the reaction doesn't involve charge transfer?

It does. n=2 comes from addition of the half-cell equations.

And what are you going to use for E0?

OP outlined this himself. I confirmed in post #3. You just convert K_sp to ΔG°, which is the same (once you take the negative) as ΔG°_cell for the cell reaction because the overall change occurring is the same in both processes (and G is path-independent), and then convert that ΔG°_cell into E°_cell.

Not sure what part of my answer you're correcting here...

[Borek]

Let's solve a different problem.

Given E₀ for Ag/Ag⁺ and K_sp for AgCl, find E for the Ag|AgCl(s) electrode immersed in the 0.01 M Cl^- solution.

[Big-Daddy]

Let's solve a different problem.

Given E₀ for Ag/Ag⁺ and K_sp for AgCl, find E for the Ag|AgCl(s) electrode immersed in the 0.01 M Cl^- solution.

Should I answer or should OP?

Spoiler:

Code: [Select]

Half-Cell Reaction: Ag(s) + Cl[sup]-[/sup](aq)  ::equil:: AgCl(s) + e[sup]-[/sup], n=1

Known Reaction Information:
Ag[sup]+[/sup](aq) + e[sup]-[/sup]  ::equil:: Ag(s), n=1 (1)
AgCl(s) ::equil:: Ag[sup]+[/sup](aq) + Cl[sup]-[/sup](aq) (2)

Half-Cell Reaction = -Reaction (2) -Reaction (1)
ΔG°[sub]cell[/sub] = -ΔG°(2)-ΔG°(1)

ΔG°(1) = -FE°(Ag/Ag[sup]+[/sup])
ΔG°(2) = -RTln(Ksp(AgCl))

so ΔG°[sub]cell[/sub] = FE°(Ag/Ag[sup]+[/sup]) + RTln(Ksp(AgCl))

E°[sub]cell[/sub] = ΔG°[sub]cell[/sub]/(-F)
E°[sub]cell[/sub] = -(RT/F)ln(Ksp(AgCl)) - E°(Ag/Ag[sup]+[/sup])

E[sub]cell[/sub] = E°[sub]cell[/sub] - (RT/F)ln(1/[Cl[sup]-[/sup](aq)])
E[sub]cell[/sub] = E°[sub]cell[/sub] + (RT/F)ln([Cl[sup]-[/sup](aq)])
E[sub]cell[/sub] = -(RT/F)ln(Ksp(AgCl)) - E°(Ag/Ag[sup]+[/sup]) + (RT/F)ln([Cl[sup]-[/sup](aq)])

E[sub]cell[/sub] = -E°(Ag/Ag[sup]+[/sup]) + (RT/F)ln([Cl[sup]-[/sup](aq)] / Ksp(AgCl))

"Cell" here really refers to half-cell

Think that's how you would do it.

I suppose your point is that the E_cell equations are the same (well, sign changed because I defined cell as an oxidation cell in Ag/AgCl(s) and reduction in Ag⁺/Ag) with and without K_sp (noting that [Cl^-(aq)] / Ksp(AgCl) = 1 / [Ag⁺(aq)], consistent with E for just Ag⁺/Ag cell if K_sp holds), that is why we can do the question faster by just finding [Ag⁺] from K_sp.

I think the problem with applying this to the OP though is that he does not know the initial E° value. That is what he needs to work out. It just so happens that ΔG°_cell is represented by the same chemical change as K_sp^-1 in the OP question.

[Borek]

I suppose your point is that the E_cell equations are the same (well, sign changed because I defined cell as an oxidation cell in Ag/AgCl(s) and reduction in Ag⁺/Ag) with and without K_sp (noting that [Cl^-(aq)] / Ksp(AgCl) = 1 / [Ag⁺(aq)], consistent with E for just Ag⁺/Ag cell if K_sp holds), that is why we can do the question faster by just finding [Ag⁺] from K_sp.

If it can be done fast, why do the detour?

I think the problem with applying this to the OP though is that he does not know the initial E° value.

I already explained why it doesn't matter. Do you need E₀ for a concentration cell?